= Solution
Let $e_k=x_k-x_*$, with $f'(x_*)=0$, $f''(x_*)\ne0$. <Taylor expansion> gives
$$
f'(x_*+e)=f''(x_*)e+\tfrac12f^{(3)}(x_*)e^2+O(e^3),\qquad f''(x_*+e)=f''(x_*)+f^{(3)}(x_*)e+O(e^2).
$$
Division in the Newton update yields
$$
\boxed{e_{k+1}=\frac{f^{(3)}(x_*)}{2f''(x_*)}e_k^2+O(e_k^3).}
$$
Thus convergence is generically quadratic when that coefficient is nonzero. With a locally Lipschitz second <derivative> and nonzero $f''(x_*)$, the weaker estimate $|e_{k+1}|\le C|e_k|^2$ also proves local <quadratic convergence>.
\b[“Quadratic at best” is false as a universal upper bound.] If the displayed coefficient vanishes, the order can be higher. For example $f(x)=x^2/2+x^4/4$ has a <local minimum>[minimum] at zero, and its <Newton root-finding iteration>[Newton optimization] update is $x_{k+1}=2x_k^3/(1+3x_k^2)$, which has <cubic convergence>. For a nondegenerate quadratic objective the method reaches the <stationary point> in one step. The expansion proves the standard generic quadratic statement while retaining and correcting the source's stronger wording explicitly.
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